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An geometry problem from China TST

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Source: China TST 4 Problem 2

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29 posts

#1 h Mar 23, 2017, 4:14 PM • 4 Y

Y by dagezjm, Adventure10, Mango247, Rounak_iitr

In $\varDelta{ABC}$ ,the excircle of $A$ is tangent to segment $BC$ ,line $AB$ and $AC$ at $E,D,F$ respectively. $EZ$ is the diameter of the circle. $B_1$ and $C_1$ are on $DF$ , and $BB_1\perp{BC}$ , $CC_1\perp{BC}$ .Line $ZB_1,ZC_1$ intersect $BC$ at $X,Y$ respectively.Line $EZ$ and line $DF$ intersect at $H$ , $ZK$ is perpendicular to $FD$ at $K$ .If $H$ is the orthocenter of $\varDelta{XYZ}$ ,prove that: $H,K,X,Y$ are concyclic.

This post has been edited 1 time. Last edited by HuangZhen, Mar 25, 2017, 9:04 AM

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#2 h Mar 23, 2017, 8:23 PM • 1 Y

Y by Adventure10

I am sorry, but is this the same $H$ ?

HuangZhen wrote:

Line $EZ$ and line $DF$ intersect at $H$

HuangZhen wrote:

$H$ is the orthocenter of $\varDelta{ABC}$

This post has been edited 1 time. Last edited by MathPanda1, Mar 23, 2017, 8:25 PM

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#4 h Mar 24, 2017, 3:17 AM • 2 Y

Y by Adventure10, Mango247

MathPanda1 wrote:

I am sorry, but is this the same $H$ ?

HuangZhen wrote:

Line $EZ$ and line $DF$ intersect at $H$

HuangZhen wrote:

$H$ is the orthocenter of $\varDelta{ABC}$

Yes,I think so...

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1135 posts

#5 h Mar 24, 2017, 3:41 AM • 2 Y

Y by Adventure10, Mango247

However, unless I read the problem wrong, the intersection of line $EZ$ and line $DF$ never coincides with the orthocenter of $ABC$ .

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4071 posts

#6 h Mar 24, 2017, 2:26 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

I don't know what the original problem is, but I can show that if we let $H$ be the orthocenter of $\triangle{XYZ}$ then the problem is true.

HuangZhen wrote:

In $\varDelta{ABC}$ ,the excircle of $A$ is tangent to segment $BC$ ,line $AB$ and $AC$ at $E,D,F$ respectively. $EZ$ is the diameter of the circle. $B_1$ and $C_1$ are on $DF$ , and $BB_1\perp{BC}$ , $CC_1\perp{BC}$ .Line $ZB_1,ZC_1$ intersect $BC$ at $X,Y$ respectively.Line $EZ$ and line $DF$ intersect at $H$ , $ZK$ is perpendicular to $FD$ at $K$ .If $H$ is the orthocenter of $\textcolor{red}{\triangle{XYZ}}$ , prove that: $H,K,X,Y$ are concyclic.

Here is my solution to the problem stated above.

Let incircle of $\triangle{ABC}$ intersect $AB,BC,CA$ at $D_2,E_2,F_2$ respectively.
Let $DF\cap BC=T$ and $D_2F_2\cap BC=T_2$
By Menelaus's, we have $\frac{BT_2}{T_2C}\times \frac{CF_2}{F_2A}\times \frac{AD_2}{D_2B}=1=\frac{BT_2}{T_2C} \times \frac{CF_2}{BD_2}=\frac{BT_2}{T_2C} \times \frac{DB}{CF}$ .
Again, by Menelaus's with $\triangle{ABC}$ and transversal $TDF$ , we have $\frac{CT}{TB}\times \frac{BD}{DA}\times \frac{AF}{FC}=1=\frac{CT}{TB}\times \frac{DB}{CF}$ .
So $\frac{BT_2}{T_2C}=\frac{CT}{TB}$ , this mean that the midpoint of $TT_2$ is $M$ , the midpoint of $BC$ .
It is well-known that $(B,C;E_2,T_2)=-1$ since $AE_2,CD_2,BF_2$ concurrent.
Reflect through $M$ give us $(B,C;E,T)=-1$ .
Pencil at point at $\infty$ of direction $\perp BC$ give us $(B_1,C_1;H,T)=-1$ and then at point $Z$ give us $(X,Y;E,T)=-1$ .
Let $XH\perp YZ=P, YH\perp XZ=Q$ , it is well-known that if $PQ\cap XY=T_3$ then $(X,Y;E,T_3)=-1$ , so $T=T_3$ .
Consider cyclic quadrilateral $XQPY$ , note that $\angle{XQY}=\angle{XPY}$ , so the quadrilateral is cyclic with center at $N$ , the midpoint of $XY$ .
By Brocard's Theorem, we get that $TH\perp ZN$ . Since $ZK\perp TH$ , we get $Z, K, N$ collinear.
Let $Z_1$ be the point defined by reflecting $Z$ through $N$ , we have $Z_1X\parallel ZY, Z_1Y\parallel ZX$ .
So $\angle{HXZ_1}=90^{\circ}=\angle{HYZ_1}$ , and we also have $\angle{HKZ_1}=90^{\circ}$
Hence $X,H,K,Y,Z_1$ lie on the same circle, in other word, $H,K,X,Y$ are concyclic and we are done.

This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Mar 24, 2017, 2:31 PM

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#7 h Mar 24, 2017, 5:30 PM • 5 Y

Y by dagezjm, freefrom_i, tenplusten, Anar24, Adventure10

Assuming $H$ is the orthocenter of $XYZ$ , I have a solution that someone should check since it seems too simple.

Let $DF$ intersect $BC$ at $P$ . Note that $(EP;BC)=-1$ , so projecting orthogonally to $BC$ onto $DF$ , we have that $(HP;B_1C_1)=-1$ . Then, we project from $Z$ onto $BC$ to conclude that $(EP;XY)=-1$ . Let $X'$ and $Y'$ be the feet of the $X$ and $Y$ altitudes in $XYZ$ . Now, if $H$ is the orthocenter of $XYZ$ , then there is an inversion about $H$ swapping $EZ,XX',YY',PK$ . Then, it suffices to show that $P$ lies on $X'Y'$ , which is obviously true since $(EP;XY)=-1$ .

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#9 h Mar 24, 2017, 8:10 PM • 3 Y

Y by tenplusten, Anar24, Adventure10

I assumed that: $H$ is orthocenter of $\triangle XYZ$

Let $W=DF\cap BC$ , since $(B,C,E,W)=-1$ projecting in $BC$ we obtain $(B_1,C_1,H,P)=-1$ . On the other hand let $X_1=XH\cap YZ$ and $Y_1=YH\cap ZX$ , since $(B_1,C_1,H,P)=-1$ we get $Z(X,Y,E,W)=-1$ , let $W'$ $=$ $X_1Y_1$ $\cap$ $XY$ $\Longrightarrow$ $Z(X,Y,E,W')=-1$ hence $W'=W$ , so from $XYX_1Y_1$ and $HKX_1Y_1$ are cyclic with diameters $AY$ and $HZ$ respectively and $W$ $=$ $XY$ $\cap$ $HK$ $\cap$ $X_1Y_1$ we get $WY.WX$ $=$ $WX_1.WY_1=WH.WK$ $\Longrightarrow$ $WX.WY=WH.WK$ hence $XYHK$ is cyclic.

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402 posts

#10 h Mar 25, 2017, 12:12 AM • 1 Y

Y by Adventure10

$\angle BDB_1=\angle CFC_1,\angle BB_1D+\angle CC_1F=180^{\circ}$ , Sine Rule tells us that $\frac {BB_1}{BD}=\frac {CC_1}{CF}\implies \frac {BB_1}{BE}=\frac {CC_1}{CE}\implies \frac {BB_1}{B_1H}=\frac {CC_1}{C_1H}$ where last equality follows from $BB_1||EH||CC_1$ , it also implies $\frac {B_1X}{ZX}=\frac {BB_1}{ZE},\frac {C_1Y}{ZY}=\frac {CC_1}{ZE}$ . Now converse of Ceva's Theorem in $\triangle ZB_1C_1$ for cevians $ZH,B_1Y,C_1X$ gives us that they are concurrent using the already proven ratio equalities at appropriate places. Hence it follows that $(B_1C_1\cap XY,E;X,Y)=-1$ . Now let $YH\cap XZ\equiv P, XH\cap YZ\equiv Q$ then $PY\perp XZ,XQ\perp YZ$ . Then again $(PQ\cap XY,E;X,Y)=-1$ . So $B_1C_1\cap XY\equiv PQ\cap XY\implies HK,PQ,XY$ are concurrent. Now $PQ$ is the radical axis of $\odot (XYPQ)$ and $\odot (ZQKHP)$ , where both circles exist due to $\angle XQY=\angle XPY=90^{\circ},\angle ZQH=\angle ZKH=\angle ZPH=90^{\circ}$ , $XY$ is the radical axis of $\odot (XHY)$ and $\odot (XYPQ)$ , if we let $l$ be the radical axis of $\odot (ZQKHP)$ and $\odot (XHY)$ then by Radical Axis theorem $XY,PQ,l$ are concurrent and $H\in l$ , then by the result that $PQ,HK,XY$ are concurrent we conclude that $l\equiv HK$ . But $K\in \odot (ZQKHP)$ hence $K$ also lies on $\odot (XHY)\implies H,K,X,Y$ are concyclic.

Note: I have also assumed $H$ to be the orthocentre of $\triangle XYZ$ . Thanks @ TheDarkIord for pointing this out.

This post has been edited 2 times. Last edited by babu2001, Mar 25, 2017, 12:17 AM
Reason: Added the note.

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#11 h Mar 25, 2017, 9:06 AM • 2 Y

Y by Adventure10, Mango247

H is the orthocenter of $\triangle {XYZ}$ ,I'm so sorry for my typing mistake.

This post has been edited 2 times. Last edited by HuangZhen, Mar 25, 2017, 9:07 AM

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#12 h Mar 25, 2017, 9:37 AM • 2 Y

Y by Adventure10, Mango247

There is something I would like to add to the problem:
$K$ lies on the median through $Z$ of $\triangle XYZ$ and its reflection in the midpoint of $XY$ lies on $\odot (XYZ)$ .

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#13 h Mar 25, 2017, 9:56 AM • 2 Y

Y by Adventure10, Mango247

In other words, K is the A-Humpty point, now aka HM point from a MR article.

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#14 h Mar 25, 2017, 2:56 PM • 2 Y

Y by Adventure10, Mango247

Use the properties of harmonic points and make an inversion about $H$ .

This post has been edited 1 time. Last edited by dagezjm, Mar 25, 2017, 2:59 PM

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#15 h Mar 26, 2020, 12:01 PM • 5 Y

Y by k12byda5h, EmPtY_sEt08, Mango247, Mango247, Mango247

Let $U$ and $V$ be the foot of $X,Y$ -altitudes of triangle $XYZ$ , and $G=BC \cap DF$ .

Note that $-1=A(BC, EG)=(B_1C_1, HG)=(XY, EG),$ so $G, U, V$ are collinear. Then $K, H, U, V$ lies on the circle with diameter $ZH$ , so by PoP we have $GK \cdot GH=GU \cdot GV=GX \cdot GY,$ or $(KHXY)$ is cyclic. We're done.

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#17 h May 19, 2024, 7:28 AM

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ZK is the median of triangle XYZ.

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